What is the physical interpretation of a field operator So far in our lecture we defined creation operators $a^{\dagger}_{n}$ in the following way, that we said: 
Somebody got you a antisymmetric or symmetric N- particle state and now $a^{\dagger}_{n}$ puts another particle in state n, so that we end up with a symmetric/antisymmetric N+1-particle state.
This interpretation is somehow clear to me in the sense that these $a^{\dagger},a$ operators avoid the cumbersome slater determinants and so on. Despite, we are still dealing with well-defined symmetrized/antisymmetrized product states that become extended or reduced by one state, which are hidden behind this notation.
Now, we also defined field-operators in QM by $\psi^{\dagger}(r) = \sum_{i;\text{all states}} \psi_i^*(r) a_i^{\dagger}.$ We said that they create a particle at position $r$. Somehow, it is not clear to me what this means: 
To create a particle at an exact position $r_0$ in QM would mean that we now have an additional state $\psi_i(r) = \delta(r-r_0)$ in our slater determinant. I doubt that this is the idea behind this. But, since the $a_i^{\dagger}$ operators act on $N$-particle state and map to $N+1$ particle states, the same must be true for $\psi^{\dagger}(r)$. Nevertheless, I have difficulties interpreting the result.
If anything is unclear, please let me know.
 A: Think of it as a change of basis. $a_i^\dagger$ creates a particle in the state $|i\rangle$. Now, this state  $|i\rangle$ can be written in terms of the position states $|r\rangle$ as 
$$ |i\rangle=\int dr\, \psi_i(r)|r\rangle,$$
thus creating a particle in this state is equivalent to create a particle in a superposition of position state with the appropriate weight $\psi_i(r)$. Equivalently, a particle localized in $|r\rangle$ can be described as being in a superposition of state 
$$|r\rangle=\sum_i \psi_i^*(r)|i\rangle,$$
and thus creating a particle in the state $|r\rangle$, the operator $\psi^\dagger(r)$ is defined by the operator $\sum_i \psi_i^*(r)\,a_i^\dagger$.
A: The $\psi_i$ in your sum do not need to be delta functions.
You can think for example as them being energy eigenfunctions
$$ \mathcal{H} \psi_i(r) = E_i \psi_i(r) $$
thus creating a particle at $r$ means that you obtain a superposition of all the possible ways a particle can be at $r$ (in this particular choise of basis):
$$ \underbrace{\psi^\dagger ( r )}_{\text{operator}} | 0 \rangle = \sum_i \overbrace{\psi_i^*(r)}^{\text{complex numbers}} | i \rangle$$
where $|0\rangle$ is the vacuum state (or ground state if you want) and $|i \rangle$ is the Fock state with one particle in the n-th mode.
You can think of this equation as stating the for each $i$, $\psi_i^*(r)$ is the probability amplitude of finding the particle at the position $r$ if you know it is in the state $i$.
